Question

Calculate the ph of a solution prepared by mixing 1.00g of imidazole (FW 6.8077) with 1.00g of imidazole hydrochloride (pKa= 6.993, FW 104.538) and diluted to 100.0 ml

Answer 1

If the strong acid HClO4 is added, the H+ will react with the base part of the buffer (C3H4N2).
moles HClO4 = M HClO4 x L HClO4 = (1.07)(0.00230) = 0.00246 moles HClO4 = 0.00247 moles H+

Moles . . . . . . .C3H4N2 + H+ ==> C3H4N2H+
Initial . . . . . . . .0.0147 . .0.00247 . . .0.00956
Change . . . . . .-0.00247 .-0.00247 . . +0.00247
Final . . . . . . . . 0.0122 . . . .0 . . . . . .0.0120

new pH = 7.08 + log (0.0122 / 0.0120) = 7.08 + 0.08 = 7.16

=======================================...

When you add H+ directly to C3H4N2 it will produce the conjugate acid C3H4N2H+ as shown above. So you would create a buffer system. The only question is, What is the base/acid ratio of the buffer to give a pH of 6.993?

pH = 7.08 + log (moles C3H4N2 / moles C3H4N2+)
6.993 = 7.08 + log (moles C3H4N2 / moles C3H4N2+)
-0.087 = (moles C3H4N2 / moles C3H4N2+)
10^-0.087 = (moles C3H4N2 / moles C3H4N2H+) = 0.818

So moles C3H4N2 = (0.818)(moles C3H4N2H+)

We also know that the total moles of C3H4N2 and C3H4N2H+ will equal the initial moles of C3H4N2 (0.0147).

moles C3H4N2 + moles C3H4N2H+ = 0.0147 . . .substituting (0.818)(moles C3H4N2H+) for moles C3H4N2 from above, we get

(0.818)(moles C3H4N2H+) + moles C3H4N2H+ = 0.0147
(1.818)(moles C3H4N2H+) = 0.0147
moles C3H4N2H+ = 0.00809

and moles C3H4N2 = 0.0147 - 0.00809 = 0.00661

Since we created 0.00809 moles of C3H4N2H+ by adding the H+ from HClO4 (which reacted with C3H4N2 in a 1:1 mole ratio),

moles H+ added = moles HClO4 = 0.00809

moles HClO4 = Molarity HClO4 x L HClO4
0.00809 = (1.07)(L HClO4)
L HClO4 = 0.00756 = 7.56 mL